Introduction

Suppose we have a collection $\cal S$ of subsets of a topological space $X$ , and for each $A\in\cal S$ we have a continuous function $f_A\colon A\to Y$ , where $Y$ is another topological space. If the functions $f_A$ agree wherever their domains intersect, then we can glue them together in the obvious way to form a new function $f\colon\cup{\cal S}\to Y$ . The theorems in this entry give some sufficient conditions for $f$ to be continuous.

Theorems

Notes

Note that the theorem for closed subsets requires the collection to be locally finite. To see that this is condition cannot be omitted, notice that any function $f\colon\R\to\R$ restricts to a continuous function on each singleton, yet need not be continuous itself.

Proofs

The two theorems are proved in essentially in the same way, but for the first theorem we need to make use of the fact that the union of a locally finite collection of closed sets is closed.

Proof of Theorem 1. Let $C$ be a closed subset of $Y$ . Then $f^{-1}(C)=\bigcup_{A\in\cal S}(A\cap f^{-1}(C)) =\bigcup_{A\in\cal S}(f|_A)^{-1}(C)$ . By continuity, each $(f|_A)^{-1}(C)$ is closed in $A$ . But by assumption each $A$ is closed in $X$ , so it follows that each $(f|_A)^{-1}(C)$ is closed in $X$ . Thus $f^{-1}(C)$ is the union of a locally finite collection of closed sets, and is therefore closed in $X$ , and so closed in $\cup\cal S$ . So $f$ is continuous.

Proof of Theorem 2. Let $U$ be an open subset of $Y$ . Then $f^{-1}(U)=\bigcup_{A\in\cal S}(A\cap f^{-1}(U)) =\bigcup_{A\in\cal S}(f|_A)^{-1}(U)$ . By continuity, each $(f|_A)^{-1}(U)$ is open in $A$ . But by assumption each $A$ is open in $X$ , so it follows that each $(f|_A)^{-1}(U)$ is open in $X$ . Thus $f^{-1}(U)$ is the union of a collection of open sets, and is therefore open in $X$ , and so open in $\cup\cal S$ . So $f$ is continuous.